Let f:N N be defined by f(n) = l 2 , ; ;if ; ;n ; ;is ; ;odd 2 , ; ; ; ; ; ; ;if ; ;n ; ;is ; ;even . for all n N. State whether the function f is bijective f is bijective. Justify your answer.

Injectivity Here, f(1) = 2 = 1, ; ; ;f(2) = 2 = 1, ; ; ;f(3) = 2 = 2, ; ; ;f(4) = 2 = 2 Thus f(2k - 1) = 2 = k ; ; ;and ; ; ;f(2k) = 2 = k f(2k - 1) = f(2k), ; ;where ; ;k N But, 2k - 1 2k, where k N f is not one-one. Surjectivity But, f is onto because range of f = N f is onto. Hence, f is not bijective.

Relations and Functions — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 1.2,Q.9, Page 11

Question

Let $f:N \to N$ be defined by $f(n) = \left\{ \begin{array}{l}\cfrac{{n + 1}}{2},\;\;if\;\;n\;\;is\;\;odd\\\cfrac{n}{2},\;\;\;\;\;\;\;if\;\;n\;\;is\;\;even\end{array} \right.$ for all $n \in N.$ State whether the function f is bijective f is bijective. Justify your answer.

Step-by-step Solution

Injectivity

Here, $f(1) = \cfrac{{1 + 1}}{2} = 1,\;\;\;f(2) = \cfrac{2}{2} = 1,\;\;\;f(3) = \cfrac{{3 + 1}}{2} = 2,\;\;\;f(4) = \cfrac{4}{2} = 2$
Thus $f(2k - 1) = \cfrac{{(2k - 1) + 1}}{2} = k\;\;\;and\;\;\;f(2k) = \cfrac{{2k}}{2} = k$

$\Rightarrow$ $f(2k - 1) = f(2k),\;\;where\;\;k \in N$

But, $2k - 1 \ne 2k,$ where $k \in N \Rightarrow f$ is not one-one.

Surjectivity

But, f is onto because range of f $=$ N

$\Rightarrow$ f is onto.

Hence, f is not bijective.

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